Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct seg {
  double l, r;
  bool operator<(const seg& other) const { return r < other.r; }
};
int n;
seg c[200005];
double totlen[200005], sy, a, b;
double rd() {
  int x;
  cin >> x;
  return x;
}
double f(double x) {
  auto it = lower_bound(c + 1, c + n + 1, seg{x, x}) - c;
  if (it == n + 1) return totlen[n];
  return totlen[it - 1] + max(x - c[it].l, 0.0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  sy = rd();
  a = rd();
  b = rd();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    c[i].l = rd();
    c[i].r = rd();
    totlen[i] = totlen[i - 1] + c[i].r - c[i].l;
  }
  int q;
  cin >> q;
  cout.precision(20);
  cout << fixed;
  while (q--) {
    double x = rd(), y = rd();
    double ux = (sy * x - a * y) / (sy - y);
    double vx = (sy * x - b * y) / (sy - y);
    double shade = f(vx) - f(ux);
    double all = vx - ux;
    cout << (b - a) * shade / all << '\n';
  }
}